Question

1. Given the impulse response, h[n duration 50 samples. (-0.9)"u[n, find the step response for a step input of h-(0.9)-10:491 -ones (1,50) s- conv(u,h) 2. Plot h and u using stem function for 50 samples only stem(10:491, s(1:50) 1. Given a system described by the following difference equation: yIn] 1143yn 1 0.4128y[n -2 0.0675x[n0.1349xn 0.675x[n-2] Determine the output y in response to zero input and the initial conditionsy-11 and yl-2] 2 for 50 samples using the following commands: a -,-1.143, 0-41281 b-[0.0675, 0.1349, 0.6751 yi- [1, 2]; %vector of İnitial condition of y in the orderly-1), y[-2], x-zeros (1,50) zi " filtic (b, a, yi ); y- filter (b, a, x, zi); %zi are the initial conditions 2. Plot the output y. stem (y) 1. Consider the LTI system described in state variable, , 0.7814 0 2 y [.9691 64493 2. Use the command ss() to produce an LTI system represented in state variable from the matrices A, B, C, D [-0.5572-0.7814;0.7814 0]; A = B- -1:0 21a c = [1.9691 6.4493]; C, -1); -1 for discrete time system ss (A, B, D, sys 1. Plot the step response using the command: figure step(sys) 2. Find the impulse response of the system using the command: figure inpulse (sys For the given step response above: . Determine the first 50 values of the step response. For the given difference equation above: . Simulate the system for a sinusoidal input x[n] = cos( n) assuming zero initial conditions. . Determine the first 100 values of the step response. For the given state variable system above: Find the time response if the system states was inosct u-0, use initialO). Simulate the system for a sinusoidal input of frequency f-0.25Hz. use command Isim().

Answer 1

Step response:

clc;close all;clear all;
h=(-0.9).^[0:49];
u=ones(1,50)
s=conv(u,h)

subplot(311)
stem([0:49],u)
xlabel('n');ylabel('u(n)');title('step input')

subplot(312)
stem([0:49],h)
xlabel('n');ylabel('h(n)');title('Impulse response h(n)')

subplot(313)
stem([0:49],s(1:50))
xlabel('n');ylabel('s(n)');title('Step response s(n)')

step input 0.8 0.6 0.4 0.2 10 40 50 20 30 Impulse response h(n) 0.5 40 20 30 50 Step response s(n) 0.8 C 0.6 D 0.4 0.2 10 40 50 20 30

Difference equation:

clc;close all;clear all;
a=[1 -1.143 0.4128]
b=[0.0675 0.1349 0.675]
%Zero input response
yi=[1 2]
x=zeros(1,50)
zi=filtic(b,a,yi)
y=filter(b,a,x,zi)
subplot(311)
stem([0:49],y(1:50))
xlabel('n');ylabel('y(n)');title('Zero input response')

%Zero state response
n=0:1:99
x=cos(pi*n/6)
y=filter(b,a,x)
subplot(312)
stem([0:99],y(1:100))
xlabel('n');ylabel('y(n)');title('Zero state response')

%Step response
x=ones(1,100)
y=filter(b,a,x)
subplot(313)
stem([0:99],y(1:100))
xlabel('n');ylabel('s(n)');title('Step response')

Zero input response 0.4 0.3 0.2 0.1 -0.1 0.2 10 40 50 20 30 Zero state response 2 -3 20 40 60 80 100 Step response 3.5 3 2.5 0.5 20 40 60 80 100

State variable description:

clc;close all;clear all;
A=[-0.5572 -0/7814 ;0.7814 0];
B=[1 -1;0 2];
C=[1.9691 6.4493];
D=0;
sys=ss(A,B,C,D,0.1)
figure;
step(sys)

figure;
impulse(sys)

figure;
x0 = [1 ; 0];
initial(sys,x0)

figure;
%Response to sinusoidal input
f=0.25
[u,t] = gensig('sin',1/f,10,0.1);
lsim(sys,[u,u],t)

Step Response 6 10 8 4 S3 4 2 2 0 0.05 0.1 0.15 0.2 0.25 0.3 0 0.05 0.1 0.15 0.2 0.25 0.3 Time [s

Impulse Response 4 10 2 -1 -2 0 0.05 0.1 0.15 0.2 0.25 0.3 0 0.05 0.1 0.15 0.2 0.25 0.3 Time [s

Response to Initial Conditions 4 2 -1 -2 0.1 0.25 0.15 Time [s] 0.2 0.3 0.05

Linear Simulation Results 10 -5 -10 2 4 8 10 Time [s]